Enthalpy of Formation: Solve CO + O_2 to CO_2

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SUMMARY

The discussion focuses on calculating the heat evolved during the combustion of carbon monoxide (CO) in the presence of excess oxygen (O2) to form carbon dioxide (CO2). The balanced chemical equation is 2CO(g) + O2(g) → 2CO2(g). The molar enthalpies provided are CO = -110.5 kJ/mol, O2 = 0.0 kJ/mol, and CO2 = -393.5 kJ/mol. The change in enthalpy (ΔH) for the reaction is calculated as ΔH = [(2)(-393.5 kJ)] - [(2)(-110.5 kJ)], and the total heat evolved for 13.39 moles of CO is determined by multiplying the ΔH by 13.39.

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ubiquinone
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Hi, I have a question involving enthalphy of formation. I tried doing this problem but I couldn't seem to get the right answer. I was wondering if anyone here may please help me out. Thanks.

Question: How much heat is evolved when 13.39 moles of CO(g) is burned in excess oxygen.
2CO(g) + O_2(g)\rightarrow 2CO_2(g)
Molar enthalpies are
CO=-110.5 kJ/mol
O_2=0.0
CO_2=-393.5 kJ/mol
 
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you figure out the Change of Enthalpy (change of H) of overall equation, and then multiply by 13.39 moles of CO to get energy (KJ).

just to make sure...

Change of H = [(2)(-110.5KJ)] -[(2)(-110.5KJ)] = ?? KJ

then mult ?? KJ by 13.39 mol

does it give you the right answer?? would like to know
 

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